How can I prove that there is no simple group of order $448=2^6\cdot 7$? I tried with Sylow's theorems, I proved that (if $G$ is simple) the number of 2-Sylows is 7 and that the number of 7-Sylows is 8 or 64, but I don't know how to continue, could you help me please?
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Let $n_2$ be the number of $2$-Sylow subgroups of $G.$ Then, $n_2$ is odd and divides $7.$ If $n_2=1$ we're done, but if $n_2=7$, by Sylow theorem, conjugation of these seven $2$-Sylow subgroups defines a homomorphism $G \to S_7.$ The kernel of this homomorphism cannot be trivial so we're done.